An object producing a pitch of 600 Hz approaches a stationary person in a straight line with a velocity of 200m/s Veocity of sound is 300 m/s The person will note a change in frequency ,as the object flies past him, equal to

Options

A. 1440 Hz
B. 240 HZ
C. 1200 Hz
D. 960 HZ

Answer

To solve the problem of the change in frequency as an object producing a pitch of 600 Hz approaches and then passes a stationary observer, we can use the Doppler effect formula. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the given values
- Frequency of the source (f) = 600 Hz
- Velocity of the source (Vs) = 200 m/s (approaching)
- Velocity of sound (V) = 300 m/s
- Velocity of the observer (Vo) = 0 m/s (stationary)

### Step 2: Calculate the frequency when the source is approaching the observer
Using the Doppler effect formula for a source moving towards a stationary observer:
\[ f' = f \left( \frac{V + Vo}{V - Vs} \right) \]
Substituting the values:
\[ f' = 600 \left( \frac{300 + 0}{300 - 200} \right) \]
\[ f' = 600 \left( \frac{300}{100} \right) \]
\[ f' = 600 \times 3 = 1800 \text{ Hz} \]

### Step 3: Calculate the frequency when the source is moving away from the observer
Now, when the source has passed the observer and is moving away, we use the Doppler effect formula again:
\[ f'' = f \left( \frac{V + Vo}{V + Vs} \right) \]
Substituting the values:
\[ f'' = 600 \left( \frac{300 + 0}{300 + 200} \right) \]
\[ f'' = 600 \left( \frac{300}{500} \right) \]
\[ f'' = 600 \times 0.6 = 360 \text{ Hz} \]

### Step 4: Calculate the change in frequency
The change in frequency (Δf) as the object flies past the observer is given by:
\[ \Delta f = f' - f'' \]
Substituting the values:
\[ \Delta f = 1800 \text{ Hz} - 360 \text{ Hz} = 1440 \text{ Hz} \]

### Final Answer
The change in frequency noted by the observer is **1440 Hz**.

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